Question

Find the moment of inertia of a cylinder of mass $M$, radius $R$ and length $L$ about an axis passing through its centre and perpendicular to its length.

Answer 1

Moment of inertia of disc $=\frac{M{R}^2}{2}$

Moment of inertia of disc parallel to its plane $=\frac{M{R}^2}{4}$ ( by perpendicular anus theorem ).

For a small part of length $dx$ amount of inertia will $=\frac{dm{R}^2}{4}+dm{x}^2$ ( by parallel axis theorem )

$\Rightarrow I={\int }_{-L/2}^{L/2}dm(\frac{R^2}{4}+x^2)$

$={\int }_{-L/2}^{+L/2}\frac{M}{L}dx(\frac{R^2}{4}+x^2)$

$={\left[\frac{M}{L}(\frac{R^2}{4}x+\frac{x^3}{3})\right]}_{-L/2}^{L/2}$

$=\frac{M}{L}(\frac{R^2}{4}L+\frac{L^3}{12})$

$=\frac{M}{4}(R^2+\frac{L^2}{3})$

Hence, the answer is $\frac{M}{4}(R^2+\frac{L^2}{3}).$